and complex exponentials with a fundamental frequency of wo=2 y(t) 1-e(3)2t1 (-3)t 点3)=∑b be-2,k otherwise b=akH(jk)→H(k)=a, for ak≠0 y(t) has only three non-zero components, therefore H(u) has a non-zero value at those three components, corresponding to A1, A2 and A, as follows H(j(0)o) bo 1 1/2 2=4 H((3)o) b3 (3.)2 4sin(3丌/2 9 j)=7丌2 H(1(-3)o) ÷2 (-3.丌/2) j(9)m2 4sin(-3丌/2) 9 4 HGw) also needs to eliminate the other frequency components of a(t)which do not exist in the output y(t) H(jk0)=0,fork≠0,± To meet the above conditions, the cutoff frequencies 1.2.3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that� � � � � and complex exponentials with a fundamental frequency of �0 = 2 . � � ej 3� t − 1 e−j 3� t 3ω 2 2 y(t) = 1 − cos t = 1 − 2 2 2 1 j(−3) � t = 1 − 2 2 1 ej(3) � t − e 2 2 1 1 j(−3)�0t = bkejk�0t = 1 − ej(3)�0t − e 2 2 k ⎧1, k = 0 − � 1 bk 2 , k = ±3 ⎧0, otherwise � bk = akH(jk�0) � H(jk�0) = bk , for ak ≤= 0. ak y(t) has only three non-zero components, therefore H(j�) has a non-zero value at those three components, corresponding to A1, A2 and A3 as follows: b0 1 H(j(0)�0) = = = 2 = A2. a0 1/2 b3 1 H(j(3)�0) = = − ÷ 2 sin(3 · ω/2) e−j(3)�/2 a3 2 j(3 · ω)2 = − j(9)ω2 ej 3� 2 4 sin(3ω/2) 9 = − j9ω2 (−j) = ω2 = A3. 4(−1) 4 1 sin(−3 · ω/2) e−j(−3)�/2 H(j(−3)�0) = b−3 = − ÷ 2 a−3 2 j(−3 · ω)2 = − 2 j(9)ω2 ej −3� 4 sin(−3ω/2) 9 = −j9ω2 (j) = ω2 = A1. 4(1) 4 H(j�) also needs to eliminate the other frequency components of x(t) which do not exist in the output y(t). � H(jk�0) = 0, for k =≤ 0, ±3 To meet the above conditions, the cutoff frequencies �1,2,3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that 6